In a single-slit Fraunhofer diffraction pattern, the first minimum occurs at angle $\theta$ satisfying:
- $a\sin\theta = \lambda/2$
- $a\sin\theta = \lambda$
- $a\sin\theta = 2\lambda$
- $a\sin\theta = n\lambda \cdot d/D$
Show answer and solution
Correct answer: B. $a\sin\theta = \lambda$
**First minimum:** $a\sin\theta = \lambda$ (slit divided into 2 halves, paths from extremes differ by $\lambda$ — pairs cancel).
Successive minima at $a\sin\theta = n\lambda$ (n = ±1, ±2, …)
central maximum is twice as wide as side fringes.